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Michael Albanese
Michael Albanese
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youYou use a local system of coordinates with x$x$-axis tangent to the boundary and y$y$ along the normal vector; let y=y(x)$y=y(x)$ the equation of the curve, so u(x,y(x))=0$u(x,y(x))=0$ (u=\phi_1$u=\phi_1$). differentiating Differentiating this equation twice at x=0$x=0$ will give u_xx(0) + y''(0)u_y(0)=0. but y''(0)=-H$u_{xx}(0) + y''(0)u_y(0)=0$, u_y=Dnubut $y''(0)=-H$, u_xx=-\lambda_1 u -Dnnu;$u_y=D_nu$, $u_{xx}=-\lambda_1 u -D_{nn}u$; so you get what you wanted.

you use a local system of coordinates with x-axis tangent to the boundary and y along the normal vector; let y=y(x) the equation of the curve, so u(x,y(x))=0 (u=\phi_1). differentiating this equation twice at x=0 will give u_xx(0) + y''(0)u_y(0)=0. but y''(0)=-H, u_y=Dnu, u_xx=-\lambda_1 u -Dnnu; so you get what you wanted.

You use a local system of coordinates with $x$-axis tangent to the boundary and $y$ along the normal vector; let $y=y(x)$ the equation of the curve, so $u(x,y(x))=0$ ($u=\phi_1$). Differentiating this equation twice at $x=0$ will give $u_{xx}(0) + y''(0)u_y(0)=0$, but $y''(0)=-H$, $u_y=D_nu$, $u_{xx}=-\lambda_1 u -D_{nn}u$; so you get what you wanted.

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user136719
user136719
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you use a local system of coordinates with x-axis tangent to the boundary and y along the normal vector; let y=y(x) the equation of the curve, so u(x,y(x))=0 (u=\phi_1). differentiating this equation twice at x=0 will give u_xx(0) + y''(0)u_y(0)=0. but y''(0)=-H, u_y=Dnu, u_xx=-\lambda_1 u -Dnnu; so you get what you wanted.